= Solution
<Conservation of energy> and <Fourier's law> give
$$
\rho c_p T_t=\frac{\partial}{\partial z}
\left(k(T)T_z\right).
$$
At steady state $kT_z=F_c$ is constant. Since $k=a/T$,
$$
\frac{d\log T}{dz}=\frac{F_c}{a}.
$$
The two temperature <boundary conditions> therefore give
$$
\boxed{T(z)=T_s\exp\left(\frac{F_c z}{a}\right)
=T_s\left(\frac{T_m}{T_s}\right)^{z/h}},
\qquad
\boxed{F_c=\frac ah\log\frac{T_m}{T_s}}.
$$
At the basal <phase boundary>, the <Stefan condition> is
$$
\rho L\dot h=F_c-F_w.
$$
The first term removes <latent heat> released by freezing, while the oceanic flux $F_w$ supplies heat to the interface. A steady shell has $\dot h=0$, hence $F_c=F_w$ and
$$
\boxed{h_\infty=\frac a{F_w}\log\frac{T_m}{T_s}}.
$$
Here $T_s$ must simultaneously satisfy the surface balance from part a with $F_c=F_w$.
Solved by gpt-5.6-sol high.
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